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First of all I would like to construct a semi formal sentence, such that the universum has at least three elements. My attempt:

$$\exists x\exists y\exists z (x\not=y\wedge y\not=z\wedge x\not=z)$$

Secondly, is there a (possibly infinite) set of sentences $T$ which has the infinite structures as models? I think it has something to do with Tarskis definition of truth. I use the following notation: $M\vDash \phi$ means $M$ satifies $\phi$, i.e the sentence $\phi$ is valid in $M$

Thirdly, how would you argue that there is no sentence $\phi$ which has the finite structures as models, I mean without a concret proof?

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  • $\begingroup$ What do you mean, "without a concrete proof"? $\endgroup$ Apr 24, 2013 at 19:21
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    $\begingroup$ Can you write a sentence saying that the universe has at least 83 objects? Do you see what this has to do with the second question? $\endgroup$ Apr 24, 2013 at 19:22
  • $\begingroup$ Well I would do it in the same way as in the first part, instead of taking $x,y,z$ I take $x_1,...,x_{83}$. I have problems seeing the relation to the second part, may you could help me. Without a concret proof I mean a simple argument which makes it somehow clear that there cant be no such sentence, but still not a correct mathematically proof $\endgroup$
    – Babla
    Apr 24, 2013 at 19:25

1 Answer 1

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  1. You've correctly written a sentence which is true only if the universe has at least three elements.

  2. Now consider the set of all sentences 'There is at least one thing', 'There are at least two things', 'There are at least three things', 'There are at least four things", etc. etc. How big would a model for all these sentences taken together have to be?

  3. The compactness theorem says that if $\Sigma$ is an inconsistent set of sentences, then there's a finite subset of $\Sigma$ which is inconsistent. This implies [the proof is given in the Wikipedia entry] that if $\varphi$ has arbitrary large finite models, it has an infinite model. So there can't be a $\varphi$ which has all finite structures (however large) as models but no infinite models. So what you need to understand here is compactness ...

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  • $\begingroup$ Thanks for your answer, I think the model has to be infintely big, correct? $\endgroup$
    – Babla
    Apr 24, 2013 at 20:10
  • $\begingroup$ For 3. Is it possible to use a similar argument which does not use compactness theorem? $\endgroup$
    – Babla
    Apr 24, 2013 at 20:19
  • $\begingroup$ If you want to avoid the compactness theorem for some reason, you could take an non-principal ultraproduct of a sequence of increasingly large models for $\Sigma$ and check that the resulting structure is infinite and is a model of $\Sigma$. ...but this requires knowing about ultraproducts, and moreover it's cheating in a way, since this is one way to prove the compactness theorem. $\endgroup$ Apr 25, 2013 at 16:41

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